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Sharon and I have finally concluded, for now, our work on the maximally dense segment problem (draft, with an errata already!), on which we have been working on and off for the past two years. Considering the algorithm itself and its derivation/proofs, I am quite happy with what we have achieved. The algorithm is rather complex, however, and it is a challenge presenting it in an accessible way. Sharon has done a great job polishing the paper, and I do hope more people would be interested in reading it and it would, eventually, inspire more work on interesting program derivations.

The basic form of the problem looks like a natural variation of the classical maximum segment sum problem: given a list of numbers, find a consecutive segment whose average, that is, sum divided by length, is maximum. The problem would be trivial without more constraints, since one could simply return the largest element, thus we usually impose a lower bound L on the length of feasible segments.

It was noticed by Huang [3], that a segment having maximum average need not be longer than 2L - 1: given a segment of 2L elements or more, we cut it in the middle. If the two halves have different averages, we keep the larger one. Otherwise the two halves have the same average. Either way, we get a shorter, feasible segment whose average is not lower. The fact hints at a trivial O(nL) algorithm: for each suffix of the list, find its best prefix upto 2L - 1 elements long.

A difficult challenge, however, is to come up with an algorithm that is O(n), independently of L. The problem can be generalised to the case where the elements do not have length 1, but each has a width, and the goal is to maximise the density — sum of the elements divided by sum of their width. It makes the problem sightly more complicated, but does not change its nature. If we go on to impose an upper bound U on the length as well, however, the problem becomes much more difficult. There was an published algorithm that claimed to be linear only to be found not so. We discovered that two later algorithms, which appeared to have concluded the problem, also fail for a boundary case. The bug is easy to fix for one of the algorithm, but might not be so for the other.

Our algorithm closely relates to that of Chung and Lu [1] and that of Goldwasser et al [2]. The algorithm is perhaps too complex to present in detail in a blog post (that’s why we need a paper!), but I will try to give an outline using pictures from the paper, my slides and poster.

One of the ways to visualise the problem is to see each element as a block, the number being the area of the block, and the density would be its height. The input is a list of (area, width) pairs, and the goal is to find a consecutive segment maximising the height. Shown below is the input list [(9,6),(6,2),(14,7),(20,4),(-10,5),(20,8),(-2,2),(27,6)], and the dashed line is their average height:

Notice that an area can be negative. In the paper, since the alphabet w is used for “window” (to be explained below), we instead refer to the width as “breadth”.

Prefixes of Suffixes, and the Window

Many optimal segment problems (finding some optimal segment of a given list) are solved by finding, for each suffix, its optimal prefix, as shown below. Each bar is a suffix of the input, and the blue part is its optimal prefix:

It is preferable that an optimal prefix of a : x can be computed from the optimal prefix of x, that is, the function computing the optimal prefix is a foldr. If it is true, the algorithm may keep a pair of (optimal segment, optimal prefix). Each time a new element is read, it computes the new optimal prefix using the previous optimal prefix, and update the optimal segment if the new prefix is better. If you like structured recursion (or the so-called “origami programming”), this form of computation is an instance of a zygomorphism.

For each optimal prefix to be computable from the previous optimal prefix, it may not extend further than the latter. We do not want the following to happen:

However, it appears to be possible for the maximally dense prefix! Imagining adding a very small, or even negative area. We might get a denser prefix by extending further to the right since the denominator is larger.

The first theorem we had to prove aimed to show that it does not matter — if a maximally dense prefix extends further than the previous one, it is going to be suboptimal anyway. Thus it is safe if we always start from the right end of the previous prefix. That is, we do not compute the maximally dense prefix of the entire input, but merely the maximally dense prefix of the previous prefix.

This is an instance of the sliding window scheme proposed by Zantema [4]. The blue part is like a “window” of the list, containing enough information to guarantee the correctness of the algorithm. As the algorithm progresses, the two ends of the window keeps sliding to the left, hence the name.

To formally show that the window contains enough information to compute the maximally dense segment, we have to clearly state what window is, and what invariant it satisfies. It turned out to be quite tricky to formally state the intuition that “the window does not always give you the optimal prefix, but it does when it matters,” and was the first challenge we met.

Since we aim at computing a segment at least L units in breadth, it might be handy to split the window into a “compulsory part” (the shortest prefix that is at least L units wide) and the rest, the “optional part”. The algorithm thus looks like this:

where the yellow bars are the compulsory parts and blue bars the optional parts. Each time we read an element into the compulsory part, zero or more elements (since the elements have non-uniform breadths) may be shifted from the compulsory part to the optional part. Then we compute a maximally dense prefix (the yellow and the blue parts together) that does not extend further than the previous one. The best among all these prefixes is the maximally dense segment.

We want a linear time algorithm, which means that all the computation from a pair of yellow-blue bars to the next pair has to be done in (amortised) constant time — how is that possible at all? To do so we will need to exploit some structure in the optional part, based on properties of density and segments.

Right-Skew Segments, and the DRSP

A non-empty list of elements x is called right-skew if, for every non-empty x₁ and x₂ such that x₁ ⧺ x₂ = x, we have density x₁ ≤ density x₂. Informally, a right-skew list is drawn as the blue wavy block below:

The rising wavy slope informally hints that the right half has a higher density than the left half wherever you make it cut. Howver, the drawing is at risk from the misunderstanding that a right-skew segment is a list of elements with ascending areas or densities. Note that neither the areas nor the densities of individual elements have to be ascending. For example, the list [(9,6),(6,2),(14,7)], with densities [1.5, 3, 2], is right-skew.

Right-skew lists are useful because of the following property. Imagining placing a list z next to x, as depicted above. To find a maximally dense prefix of z ⧺ x starting with z, it is sufficient to consider only z and z ⧺ x — nothing in the middle, such as z ⧺ x₁, can be denser than the two ends!

Given a window with compulsory part c and optional part x, if we can partition x into x₁ ⧺ x₂ ⧺ ... ⧺ xn, such that x₁, x₂, … xn are all right-skew, then to compute the maximally dense prefix of c ⧺ x, we only need to consider c, c ⧺ x₁, c ⧺ x₁ ⧺ x₂,… and c ⧺ x₁ ⧺ x₂ ⧺ ... ⧺ xn.

Such a partition is always possible for any list x — after all, each element itself constitute a singleton right-skew list. However, there is one unique right-skew partition such that the densities of x₁, x₂, … xn are strictly decreasing. This is called the decreasing right-skew partition (DRSP) of x. We will partition the optional part of the window into its DRSP. A window now looks like the picture below:

Sharon summarised many nice properties of DRSP in the paper, for which we unfortunately do not have space here. We will only look at some properties that matters for this blog post. Firstly, consider the diagram below:

In the bottom row, the leftmost block is the density of c, and the second block is the density of c ⧺ x₁, etc. If segments x₁, x₂, … xn have decreasing densities, the densities of c, c ⧺ x₁, c ⧺ x₁ ⧺ x₂,… and c ⧺ x₁ ⧺ x₂ ⧺ ... ⧺ xn must be bitonic — first ascending, then descending. It helps to efficiently locate the maximally dense prefix.

Secondly, the DRSP can be built and maintained in a foldr. The following diagram depicts how the DRSP for the list of areas [1,4,2,5,3] (all with breadth 1) can be built by adding elements from the left one by one (which eventually results in one big partition):

The rule is that blocks newly added from the left keeps merging with blocks to its right until it encounters a block shorter than itself. The top-left of the diagram indicates that the DRSP of (3 is itself. Since 5 > 3, adding 1 results in a partition containing two segments. When 2 is added, it is merged with 5 to form a new segment with density 3.5. No merging is triggered with the addition of 4 since 4 > 3.5 and thus [4,3.5,3] form a decreasing sequence. Newly added 1 first merges 4, forming a block having density 2.5. Since 2.5 < 3.5, it again merges with the block [2,5]. Eventually all elements are grouped into one segment with density 3. One important thing here is that adding a new element only involves merging some initial parts of the DRSP.

Algorithm Overview

Recall that our algorithm computes, for each suffix, a prefix (a window) that is possibly optimal and contains enough information to compute all future optimal solutions. Since a feasible prefix is wider than L, we split it into a (yellow) compulsory part and a (blue) optional part. To obtain a linear time algorithm, we have to compute one row from the previous row in amortised constant time (the corresponding diagram is duplicated here):

The diagram below depicts how to go from one row to the next. The blue part is partitioned into DRSP. Each time an element is added to the yellow part, some elements may be shifted to the blue part, and that may trigger some right-skew segments in the blue part to be merged (second row). Then we look for a maximally dense prefix by going from right to left, chopping away segments, until we find the peak (third row):

Note that the operation shown on the third row (chopping to find the maximum) always chop away a right-skew segment in its entirety. It is important that the merging happens at the left end of the optional part, while the chopping happens at the right end. By using a tree-like data structure, each merging can be a O(1) operation. With the data structure, we may argue that, since each element can be merged at most once, throughout the algorithm only O(n) merging could happen. Similarly, each element can be chopped away at most once, so the chopping could happen at most O(n) time as well. Therefore the operations in the second and third rows above are both amortised O(1).

Problem with Having an Upper Bound

The discussion so far already allows us to develop an algorithm for the maximally dense segment problem without an upper bound on the breadth of feasible segments. Having the upper bound makes the problem much harder because, different from the chopping depicted above, an upper bound may cut through a right-skew segment in the middle:

And a right-skew segment, with some elements removed, might not be right-skew anymore!

Our solution is to develop another data structure that allows efficient removal from the right end of a DRSP, while maintaining the DRSP structure. The final configuration of a window looks like the diagram below, where the new data structure is represented by the green blocks:

Unfortunately, it is inefficient to add new elements from the left into the green blocks. Therefore we have to maintain the window in a way similar to how a queue is implemented using two lists. New elements are added from the left into the blue blocks; when we need to remove element from the right of a block, it is converted to a green block in large chunks.

Page 3: “This input sequence does not have a solution…” what we meant was “This input does not have a prefix that is within bounds.” We used another example where the input does not have a feasible segment at all before changing to example, but I forgot to change the text accordingly.

The problem of finding a maximally dense segment (MDS) of a list is a generalisation of the well-known maximum segment sum (MSS) problem, but its solution is more challenging. We extend and illuminate some recent work on this problem with a formal development of a linear-time online algorithm, in the form of a sliding window zygomorphism. The development highlights some elegant properties of densities, involving partitions which are decreasing and all right-skew.

In a previous paper of mine, regrettably, I wrongly attributed the origin of the maximum segment sum problem to Dijkstra and Feijen’s Een methode van programmeren. In fact, the story behind the problem was told very well in Jon Bentley’s Programming Pearls.

The Problem, and the Linear-Time Algorithm

Given a list of numbers, the task is to compute the largest possible sum of a consecutive segment. In a functional language the problem can be specified by:

mss = max . map sum . segments

where segments = concat . map inits . tails enlists all segments of the input list, map sum computes the sum of each of the segments, before max :: Ord a ⇒ [a] → a picks the maximum. The specification, if executed, is a cubic time algorithm. Yet there is a linear time algorithm scanning through the list only once:

Both the specification and the linear time program are short. The program is merely a foldr that can be implemented as a simple for-loop in an imperative language. Without some reasoning, however, it is not that trivial to see why the program is correct (hint: the foldr computes a pair of numbers, the first one being the maximum sum of all prefixes of the given list, while the second is the maximum sum of all segments). Derivation of the program (given below) is mostly mechanical, once you learn the basic principles of program calculation. Thus the problem has become a popular choice as the first non-trivial example of program derivation.

Origin

Jon Bentley recorded in Programming Pearls that the problem was proposed by Ulf Grenander of Brown University. In a pattern-matching procedure he designed, a subarray having maximum sum is the most likely to yield a certain pattern in a digitised image. The two dimensional problem took too much time to solve, so he simplified to one dimension in order to to understand its structure.

In 1977 [Grenander] described the problem to Michael Shamos of UNILOGIC, Ltd. (then of Carnegie-Mellon University) who overnight designed Algorithm 3. When Shamos showed me the problem shortly thereafter, we thought that it was probably the best possible; … A few days later Shamos described the problem and its history at a Carnegie-Mellon seminar attended by statistician Jay Kadane, who designed Algorithm 4 within a minute.

Jon Bentley, Programming Pearls (1st edition), page 76.

Jay Kadane’s Algorithm 4 is the now well-known linear time algorithm, the imperative version of the functional program above:

Algorithm 3, on the other hand, is a divide and conquer algorithm. An array a is split into two halves a₁ ⧺ a₂, and the algorithm recursively computes the maximum segment sums of a₁ and a₂. However, there could be some segment across a₁ and a₂ that yields a good sum, therefore the algorithm performs two additional loops respectively computing the maximum suffix sum of a₁ and the maximum prefix sum of a₂, whose sum is the maximum sum of segment crossing the edge. The algorithm runs in O(N log N) time. (My pseudo Python translation of the algorithm is given below.)

In retrospect, Shamos did not have to compute the maximum prefix and suffix sums in two loops each time. The recursive function could have computed a triple quadruple of (maximum prefix sum, maximum segment sum, maximum suffix sum, and sum of the whole array) for each array. The prefix and suffix sums could thus be computed bottom-up. I believe that would result in a O(N) algorithm. This linear time complexity might suggest that the “divide” is superficial — we do not have to divide the array in the middle. It is actually easier to divide the array into a head and a tail — which was perhaps how Kadane quickly came up with Algorithm 4!

A Functional Derivation

I learned the function derivation of the maximum segment sum problem from one of Jeremy’s papers [3] and was very amazed. It was perhaps one of the early incident that inspired my interest in program calculation. The derivation does not appear to be very well known outside the program derivation circle — not even for functional programmers, so I would like to redo it here.

The purpose of the book-keeping transformation above is to push max . map sum closer to inits. The fragment max . map sum . inits is a function which, given a list of numbers, computes the maximum sum among all its prefixes. We denote it by mps, for maximum prefix sum. The specification has been transformed to:

mss = max . map mps . tails

This is a common strategy for segment problems: to solve a problem looking for an optimal segment, proceed by looking for an optimal prefix of each suffix. (Symmetrically we could process the list the other way round, look for an optimal suffix for each prefix.)

We wish that mps for each of the suffixes can be efficiently computed in an incremental manner. For example, to compute mps [-1,3,3,-4], rather than actually enumerating all suffixes, we wish that it can be computed from -1 and mps [3,3,-4] = 6, which can in turn be computed from 3 and mps [3,-4] = 3, all in constant time. In other words, we wish that mps is a foldr using a constant time step function. If this is true, one can imagine that we could efficiently implement map mps . tails in linear time. Indeed, scanr f e = map (foldr f e) . tails!

The aim now is to turn mps = max . map sum . inits into a foldr. Luckily, inits is actually a foldr. In the following we will perform foldr-fusion twice, respectively fusing map sum and max into inits, thus turning the entire expression into a foldr.

In words, mps (x : xs), the best prefix sum of x : xs, can be computed by zmax x (mps xs). The definition of zmax says that if x + mps xs is positive, it is the maximum prefix sum; otherwise we return 0, sum of the empty prefix.
Therefore, mss can be computed by a scanr:

Many functional derivations usually stop here. This gives us an algorithm that runs in linear time, but takes linear space. A tupling transformation eliminates the need for linear space:

mss = snd . (head &&& max) . scanr zmax 0

where (f &&& g) a = (f a, g a). The part (head &&& max) . scanr zmax 0 returns a pair, the first component being the result of mps, the second mss. By some mechanical simplification we get the final algorithm:

A Relational Derivation?

The maximum segment sum problem later turned out to be a example of Richard and Oege’s Greedy Theorem [2]. It is an exercise in the Algebra of Programming book, but I have not seen the solution given anywhere. For completeness, I recorded a relational derivation in a paper of mine about some other variations of the maximum segment sum problem[4].

Page 3: “This input sequence does not have a solution…” what we meant was “This input does not have a prefix that is within bounds.” We used another example where the input does not have a feasible segment at all before changing to example, but I forgot to change the text accordingly.

Expository Code

The expository program [download here] intends to be an executable implementation of the code in the paper. For clarity we use Haskell lists for all sequences, and do not implement optimisations such as storing the areas and breadths of segments, thus the program is not linear time yet. A linear time implementation will follow soon.

The code is mostly consistent with the paper, with some minor differences: many functions take an additional argument of type BreadthSpec = (Breadth, Maybe Breadth). The first component is the lower bound, while the second component is an optional upperbound. The main function is:

mds :: BreadthSpec -> [Elem] -> Maybe [Elem]

which takes a BreadthSpec and switches between the modes with or without an upper bound.

To try the code, you may either load the module Main into ghci and invoke the function mds:

mds (lb, Just ub) [x1, x2, x3, ... ]

or load the module Test and run some QuickCheck properties:

Test.QuickCheck.quickCheck (prop_mds_correct bb (lb, Just ub))

where lb and ub are the lower and upper bounds, and bb is the bound on breadths of generated test elements. The property prop_mds_correct asserts that mds (lb,ub) x =d mds_spec (lb,ub) x for all x.

Many types having area, breadth, and density defined are collected into a type class Block, and functions like maxChop are define on the type class.

DRSP: specification of right-skew segments and DRSP, with functions including rightSkew, sdars, lrsp, and drsp.

DPTrees: defining DTrees and PTrees, and functions like addD and prependD allowing one to construct DRSPs in a fold.

Utilities: some utility functions processing lists.

Test: a module defining some QuickCheck properties to test the code.

Linear Time Implementation

A linear time implementation can be downloaded here. The program uses Data.Sequence to represent the compulsory part and the first level of the DForest and the PForest of the window, as well as annotating them with areas and breadths. The subtrees of a DTree and a PTree, however, can be represented simply by snoc-lists and cons-lists respectively.

Organisation of the code is the same as the first program.

Proofs

Proofs accompanying the paper [PDF]. Theorems and lemmas are labelled with both their own numbers as well as the numbers in the paper, if any. For example, Lemma A.1 (3.1) is Lemma 3.1 in the paper.

If you think you know everything you need to know about binary search, but have not read Netty van Gasteren and Wim Feijen’s note The Binary Search Revisited, you should.

In the FLOLAC summer school this year we plan to teach the students some basics of Hoare logic and Dijkstra’s weakest precondition calculus — a topic surprisingly little known among Taiwanese students. Binary search seems to be a topic I should cover. A precise specification will be given later, but basically the requirement is like this: given a sorted array of N numbers and a value to search for, either locate the position where the value resides in the array, or report that the value does not present in the array, in O(log N) time.

Given that everybody should have learned about binary search in their algorithm class, you would not expect it to be a hard programming task. Yet in his popular book Programming Pearls, Jon Bentley noted that surprisingly few professions programmers managed to implement the algorithm without bugs at their first attempt. I tried it myself and, being very careful (and having learned something about program construction already), I produced a program which I believe to be correct, but rather inelegant. I tried it on my students and they did produce code with some typical bugs. So it seems to be a good example for a class about correct program construction.

The Van Gasteren-Feijen Approach

The surprising fact that van Gasteren and Feijen pointed out was that binary search does not apply only to sorted lists! In fact, the usual practice comparing binary search to searching for a word in a dictionary is, according to van Gasteren and Feijen, a major educational blunder.

Van Gasteren and Feijen considered solving a more general problem: let M and N be integral numbers with M < N, and let Φ be a relation such that M Φ N (where Φ is written infix), with some additional constraints to be given later. The task is to find l such that

Notice first that the loop guard l+1 ≠ r, if satisfied, guarantees that l and r are not adjacent numbers, therefore assigning m := (l + r) / 2 establishes l < m < r, and thus the bound r - l is guaranteed to decrease. The if statement clearly maintains the invariant, if at least one of the guards are always satisfied:

l Φ r ∧ l < m < r ⇒ l Φ m ∨ m Φ r(*)

For Φ satisfying the condition above, at the end of the loop we will find some l such that l Φ (l+1).

What relations satisfy (*)? Examples given by van Gasteren and Feijen include:

i Φ j = a[i] ≠ a[j] for some array a. Van Gasteren and Feijen suggested using this as the example when introducing binary search.

i Φ j = a[i] < a[j],

i Φ j = a[i] × a[j] ≤ 0,

i Φ j = a[i] ∨ a[j], etc.

Searching for a Key in a Sorted Array

To search for a key K in an ascending-sorted array a, it seems that we could just pick this Φ:

i Φ j = a[i] ≤ K < a[j]

and check whether a[i] = K after the loop. There is only one problem, however -- we are not sure we can establish the precondition a[l] ≤ K < a[r]!

Van Gasteren and Feijen's solution is to add to imaginary elements to the two ends of the array. That is, for a possibly empty array a[0..N), we imagine two elements a[-1] such that a[-1] ≤ x and a[N] such that x < a[N] for all x. I believe this is equivalent to using this Φ:

Do not worry about the idea "adding" elements to a. The invariant implies that -1 < m < N, thus a[-1] and a[N] are never accessed, and the array a needs not be actually altered. They are just there to justify the correctness of the program. It also enables us to handle possibly empty arrays, while the loop body seems to be designed for the case when the range [l..r) is non-empty.

I would like to be able to derive this program in class, since this appears to be the more popular version. Apart from the presence of break, which I do not yet know of a easy variation of Hoare logic that helps to derive it, to relate the test a[m] < K to l := m + 1 I will have to bring in the fact that a is sorted in an earlier stage of the development. Thus it is harder to put it in a more general picture.

For several reasons I used to believe that Bentley's program could be preferred, for example, it seems to shrink the range more effectively, assigning l and r to m + 1 and m - 1, rather than m. On a second thought I realised that it might not be true. Variable l can be assigned m + 1 because the possibility of a[m] = K is covered in another case with an early exit, and r is assigned m - 1 because this algorithm represents an array segment with an inclusive right bound, as opposed to the previous algorithm.

The two algorithms do not solve exactly the same problem. With multiple occurrences of K in the array, Bentley's algorithm is non-deterministic about which index it returns, while the van Gasteren-Feijen algorithm, enforced by the specification, always returns the largest index. When K does not appear in the array, van Gasteren and Feijen's program could be more efficient because it needs only one comparison in the loop, rather than two as in Bentley's case (I am assuming that the last comparison is a catch-all case and need not be implemented). What if K does present in the array? An analysis by Timothy J. Rolfe concluded that a single-comparison approach is still preferable in average -- benefit of the early exit does not outweigh the cost of the extra comparison in the loop.

On Computing the Middle Index

There are some other interesting stories regarding the assignment m := (l + r) / 2. Joshua Bloch from Google noted that for large arrays, adding l and r may cause an overflow, and Bloch was not picking on Bentley -- the bug was reported by Sun. Bloch suggests one of the following:

Since the publication of the blog post, there have been numerous discussions on whether it should be considered a bug in the binary search algorithm or the integer datatype, and some more machine dependent issues like whether one may have an array so large that cannot be indexed by an int, etc.

Exercise?

Among the exercises suggested by van Gasteren and Feijen, this one caught my eye: let array a[0,N), with 0 < N, be the concatenation of a strictly increasing and a strictly decreasing array. Use binary search to find the maximum element. (For this problem I think it is reasonable to assume that the two sub-arrays could be empty, while a is non-empty.) This happens to a sub-routine I needed for an O(N log N) algorithm for the maximum segment density problem (there are linear-time algorithms for this problem, though), and I do remember I started off treating it as an unimportant sub-routine but had a hard time getting it right. I am glad that now I know more about it.

Yu-Han Lyu and I were studying some paper from the algorithm community, and we noticed a peculiar kind of argument. For a much simplified version, let X and D be two relations of type A → B, denoting two alternative approaches to non-deterministically compute possible solution candidates to a problem. Also let ≤ be a transitive relation on B, and ≥ its converse. The relation min ≤ : {B} → B, given a set, returns one of its elements that is no larger (under ≤) than any elements in the set, if such a minimum exists.
We would like find solution as small as possible under ≤.

When arguing for the correctness of its algorithm, the paper we are studying claims that the method X is no worse than D in the following sense: if every solution returned by D is no better than some solution returned by X, which we translate to:

D ⊆ ≥ . X

then the best (smallest) solution by X must be no worse than (one of the) best solutions returned by D:

min ≤ . ΛX ⊆ ≤ . min ≤ . ΛD

where Λ converts a relation A → B to a function A → {B} by collecting its results to a set. Note that, awkwardly, X and D are swapped to different sides of relational inclusion.

“What? How could this be true?” was my first reaction. I bombarded Yu-Han with lots of emails, making sure that we didn’t misinterpret the paper. An informal way to see it is that since every result of D is outperformed by something returned by X, collectively, the best result among the latter must is “lower-bounded” by the optimal result of D. But this sounds unconvincing to me. Something is missing.

Totality and Well-Boundedness

It turns out that the reasoning can be correct, but we need some more constraints on D and ≤. Firstly, D must yield some result whenever X does. Otherwise it could be that D ⊆ ≥ . X is true but ΛD returns an empty set, while ΛX still returns something. This is bad because X is no more a safe alternative of D — it could sometimes do too much. One way to prevent it from happening so is to demand that ΛD = dom ∈ . ΛD, where ∈ is the membership relation, and dom ∈, the domain of ∈, consists only of non-empty sets. It will be proved later that this is equivalent to demanding that D be total.

Secondly, we need to be sure that every non-empty set has a minimum, or min ≤ always yields something for non-empty sets. Therefore min ≤ . ΛD would not fall back to the empty relation. Formally, it can be expressed as dom ∈ = dom (min ≤). Bird and de Moor called this property well-boundedness of ≤.

Recall that min ≤ = ∈ ∩ ≤/∋. The part ∈ guarantees that min ≤ returns something that is in the given set, while ≤/∋ guarantees that the returned value is a lower-bound of the given set. Since ΛD (as well as ΛX) is a function, we also have min ≤ . ΛD = D ∩ ≤/D°, following from the laws of division.

Later we will prove an auxiliary lemma stating that if ≤ is well-bounded, we have:

≤/∋ . dom ∈ ⊆ ≤ . min ≤ . dom ∈

The right-hand side, given a non-empty list, takes its minimum and returns something possibly smaller. The left-hand side merely returns some lower-bound of the given set. It sounds weaker because it does not demand that the set has a minimum. Nevertheless, the inclusion holds if ≤ is well-bounded.

An algebraic proof of the auxiliary lemma was given by Akimasa Morihata. The proof, to be discussed later, is quite interesting to me because it makes an unusual use of indirect equality. With the lemma, proof of the main result becomes rather routine:

Proof Using Enriched Indirect Equality

Now we have got to prove that ≤/∋ . dom ∈ ⊆ ≤ . min ≤ . dom ∈ provided that ≤ is well-bounded. To prove this lemma I had to resort to first-order logic. I passed the problem to Akimasa Morihata and he quickly came up with a proof. We start with some preparation:

The second unusual aspect is that rather than starting from one of X ⊆ ≤/(min ≤)° . dom ∈ or X ⊆ ≤ . min ≤ . dom ∈ and ending at another, Morihata’s proof took the goal as a whole and used rules like (P ⇒ Q) ⇒ (P ⇒ P ∧ Q). The proof goes:

The steep list problem is an example we often use when we talk about tupling. A list of numbers is called steep if each element is larger than the sum of elements to its right. Formally,

steep [] = True
steep (x : xs) = x > sum xs ∧ steep xs

The definition above is a quadratic-time algorithm. Can we determine the steepness of a list in linear time? To do so, we realise that we have to compute some more information. Let steepsum = ⟨steep, sum⟩ where ⟨f, g⟩ x = (f x, g x), we discover that steepsum can be computed as a foldr:

In the Software Construction course we also talked about list homomorphism, that is, functions that can be defined in the form

h [] = e
h [a] = f a
h (xs ⧺ ys) = h xs ⊚ h ys

where ⊚ is associative and e is the identity element of ⊚. The discussion would be incomplete if we didn’t mention the third homomorphism theorem: if a function on lists can be computed both from right to left and from left to right, that it, both by a foldr and a foldl, it can be computed from anywhere in the middle by a homomorphism, which has the potential of being parallelised. Hu sensei had this idea using steepsum as an exercise: can we express steepsum as a foldl, and a list homomorphism?

Unfortunately, we cannot — steepsum is not a foldl. To determining the steepness from left to right, we again have to compute some more information — not necessarily in the form of a tuple.

Right Capacity

The first idea would be to tuple steepsum with yet another element, some information that would allow us to determine what could be appended to the right of the input. Given a list of numbers xs, let rcap xs (for right capacity) be an (non-inclusive) upperbound: a number y < rcap xs can be safely appended to the right of xs without invalidating the steepness within xs. It can be defined by:

rcap [] = ∞
rcap (x : xs) = (x - sum xs) ↓ rcap xs

where ↓ stands for binary minimum. For an explanation of the second clause, consider x : xs ⧺ [y]. To make it a steep list, y has to be smaller than x - sum xs and, furthermore, y has to be small enough so that xs ⧺ [y] is steep. For example, rcap [9,5] is 4, therefore [9,5,3] is still a steep list.

Following the theme of tupling, one could perhaps try to construct ⟨steep, sum, rcap⟩ as a foldl. By doing so, however, one would soon discover that rcap itself contains all information we need. Firstly, a list xs is steep if and only if rcap xs is greater than 0:Lemma 1: steep xs ⇔ rcap xs > 0.Proof: induction on xs.case[]: True ⇔ ∞ > 0.case(x : xs):

∎
Therefore we have rcap = foldl (λ y x → (y - x) ↓ x) ∞. If the aim were to determine steepness from left to right, our job would be done.

However, the aim is to determine steepness from both directions. To efficiently compute rcap using a foldr, we still need to tuple rcap with sum. In summary, the function ⟨sum, rcap⟩ can be computed both in terms of a foldl:

List Homomorphism

Abbreviate ⟨sum, rcap⟩ to sumcap. The aim now is to construct ⊚ such that sumcap (xs ⧺ ys) = sumcap xs ⊚ sumcap ys. The paper Automatic Inversion Generates Divide-and-Conquer Parallel Programs by Morita et al. suggested the following approach (which I discussed, well, using more confusing notations, in a previous post): find a weak inverse of ⟨sum, rcap⟩, that is, some g such that sumcap (g z) = z for all z in the range of sumcap. Then we may take z ⊚ w = sumcap (g z ⧺ g w).

For this problem, however, I find it hard to look for the right g. Anyway, this is the homomorphism that seems to work:

In the previous post we reviewed Hans Zantema’s algorithm for solving longest segment problems with suffix and overlap-closed predicates. For predicates that are not overlap-closed, Zantema derived a so-called “windowing” technique, which will be the topic of this post.

A brief review: the longest segment problem takes the form:

max# ∘ p ◁ ∘ segs

where segs :: [a] → [[a]], defined by segs = concat ∘ map inits ∘ tails returns all consecutive segments of the input list; p ◁ is an abbreviation for filter p, and max# :: [[a]] → [a] returns the longest list from the input list of lists. In words, the task is to compute the longest consecutive segment of the input that satisfies predicate p.

A predicate p is suffix-closed if p (xs ⧺ ys) ⇒ p ys. For suffix-closed p, Zantema proposed a technique that, from a high-level point of view, looks just like the right solution to such problems. We scan through the input list using a foldr from the right to the left, during which we try to maintain the longest segment satisfying p so far. Also, we keep a prefix of the list that is as long as the currently longest segment, which we call the window.

If, when we move one element to the right, the window (now one element longer than the currently longest segment) happens to satisfy p, it becomes the new optimal solution. Otherwise we drop the right-most element of the window so that it slides leftwards, retaining the length. Notice that it implies that we’d better represent the window using a queue, so we can efficiently add elements from the left and drop from the right.

Derivation of the algorithm is a typical case of tupling.

Tupling

Given a function h, we attempt to compute it efficiently by turning it into a foldr. It would be possible if the value of the inductive case h (x : xs) were determined solely by x and h xs, that is:

h (x : xs) = f x (h xs)

for some f. With some investigation, however, it would turn out that h (x : xs) also depends on some g:

h (x : xs) = f x (g (x : xs)) (h xs)

Therefore, we instead try to construct their split⟨ h , g ⟩ as a fold, where the split is defined by:

⟨ h , g ⟩ xs = (h xs, g xs)

and h = fst . ⟨ h , g ⟩.

If ⟨ h , g ⟩ is indeed a fold, it should scan through the list and construct a pair of a h-value and a g-value. To make it feasible, it is then hoped that g (x : xs) can be determined by g xs and h xs. Otherwise, we may have to repeat the process again, making the fold return a triple.

Segment/Prefix Decomposition

Let us look into the longest segment problem. For suffix-closed p it is reasonable to assume that p [] is true — otherwise p would be false everywhere. Therefore, for the base case we have max# ∘ p ◁ ∘ segs ▪ [] = []. We denote function application by ▪ to avoid too many parentheses.

Now the inductive case. It is not hard to derive an alternative definition of segs:

It suggests that we maintain, by a foldr, a pair containing the longest segment and the longest prefix satisfying p (that is, max# ∘ p ◁ ∘ inits). It is then hoped that max# ∘ p ◁ ∘ inits ▪ (x : xs) can be computed using max# ∘ p ◁ ∘ inits ▪ xs. And luckily, it is indeed the case, implied by the following proposition proved in an earlier post:

Proposition 1: If p is suffix-closed, we have:

p ◁ ∘ inits ▪ (x : xs) = finits (max# ∘ p ◁ ∘ inits ▪ xs)

where finits ys = p ◁ ∘ inits ▪ (x : ys).

Proposition 1 says that the list (or set) of all the prefixes of x : xs that satisfies p can be computed by the longest prefix of xs (call it ys) satisfying p, provided that p is suffix-closed. A naive way to do so is simply by computing all the prefixes of x : ys and do the filtering again, as is done in finits.

This was the route taken in the previous post. It would turn out, however, to come up with an efficient implementation of f we need some more properties from p, such as that it is also overlap-closed.

The “Window”

Proposition 1 can be strengthened: to compute all the prefixes of x : xs that satisfies p using finits we do not strictly have to start with ys. Any prefix of xs longer than ys will do.

In particular, we may choose i to be the length of the longest segment:

Lemma 1:

length ∘ max# ∘ p ◁ ∘ segs ▪ xs ≥
length ∘ max# ∘ p ◁ ∘ inits ▪ xs

Appealing to intuition, Lemma 1 is true because segs xs is a superset of inits xs.

Remark: Zantema proved Proposition 1 by contradiction. The purpose of an earlier post was to give a constructive proof of Proposition 1, which was considerably harder than I expected. I’d be interested to see a constructive proof of Proposition 2.

Maintaining the Longest Segment and the Window

The task now is to express ⟨ max# ∘ p ◁ ∘ segs , window ⟩ as a foldr. We can do so only if both max# ∘ p ◁ ∘ segs ▪ (x : xs) and window (x : xs) can be determined by max# ∘ p ◁ ∘ segs ▪ xs and window xs. Let us see whether it is the case.

The Algorithm

As is typical of many program derivations, after much work we deliver an algorithm that appears to be rather simple. The key invariants that made the algorithm correct, such as that s is the optimal segment and that w is as long as s, are all left implicit. It would be hard to prove the correctness of the algorithm without these clues.

We are not quite done yet. The window w had better be implemented using a queue, so that init w can be performed efficiently. The algorithm then runs in time linear to the length of the input list, provided that p (x : w) can be performed in constant time -- which is usually not the case for interesting predicates. We may then again tuple the fold with some information that helps to compute p efficiently. But I shall stop here.

I spent most of the week preparing for the lecture on Monday, in which we plan to talk about segment problems. One of the things we would like to do is to translate the derivations in Hans Zantema’s Longest Segment Problems to Bird-Meertens style. Here is a summary of the part I managed to do this week.

Zantema’s paper considered problems of this form:

max# ∘ p ◁ ∘ segs

where segs :: [a] → [[a]], defined by segs = concat ∘ map inits ∘ tails returns all consecutive segments of the input list; p ◁ is a shorter notation for filter p, and max# :: [[a]] → [a] returns the longest list from the input list of lists. In words, the task is to compute the longest consecutive segment of the input that satisfies predicate p.

Of course, we have to assume certain properties from the predicate p. A predicate p is:

For example, ascending is suffix and overlapping-closed, while p xs = (all (0 ≤) xs) ∧ (sum xs ≤ C) for some constant C is suffix-closed but not overlap-closed. Note that for suffix-closed p, it is reasonable to assume that p [] is true, otherwise p would be false everywhere. It also saves us some trouble being sure that max# is always applied to a non-empty set.

Like what we do with the classical maximum segment sum, if we can somehow turn max# ∘ p ◁ ∘ inits into a fold, we can then implement map (foldr ...) ∘ tails by a scanr. Let us denote max# ∘ p ◁ ∘ inits by mpi.

If you believe in structural recursion, you may attempt to fuse map# ∘ p ◁ into inits by fold-fusion. Unfortunately, it does not work this way! In the fold-fusion theorem:

h ∘ foldr f e = foldr g (h e) ⇐ h (f x y) = g x (h y)

notice that x and y are universally quantified, which is too strong for this case. Many properties we have, to be shown later, do need information from the context — e.g. some properties are true only if y is the result of inits.

Trimming Unneeded Prefixes

One of the crucial properties we need is the following:

Proposition 1: If p is suffix-closed, we have:

p ◁ ∘ inits ▪ (x : xs) =
p ◁ ∘ inits ▪ (x : max# ∘ p ◁ ∘ inits ▪ xs)

For some intuition, let x = 1 and xs = [2,3,4]. The left-hand side first compute all prefixes of xs:

[] [2] [2,3] [2,3,4]

before filtering them. Let us assume that only [] and [2,3] pass the check p. We then pick the longest one, [2,3], cons it with 1, and compute all its prefixes:

[] [1] [1,2] [1,2,3]

before filtering them with p again.

The right-hand side, on the other hand, performs filtering on all prefixes of [1,2,3,4]. However, the proposition says that it is the same as the left-hand side — filtering on the prefixes of [1,2,3] only. We lose nothing if we drop [1,2,3,4]. Indeed, since p is suffix-closed, if p [1,2,3,4] were true, p [2,3,4] would have been true on the right-hand side.

Proof of Proposition 1 was the topic of a previous post. The proposition is useful for us because:

with the definition of ⊙ given above. It turns out to be a rather simple algorithm: we scan through the list, and in each step we choose among three outcomes: [], [x], and x : ys. Like the maximum segment sum problem, it is a simple algorithm whose correctness is that that easy to justify.

The algorithm would be linear-time if ⊙ can be computed in constant-time. With the presence of p in ⊙, however, it is unlikely the case.

Efficient Testing

So let us compute, during the fold, something that allows p to be determined efficiently. Assume that there exists some φ :: [A] → B that is a fold (φ = foldr ⊗ ι for some ⊗ and ι), such that p (x : xs) = p xs ∧ f x (φ xs) for some f. Some example choices of φ and f:

Notice that the property f = fst ∘ ⟨ f , g ⟩ is true when the domain of f is in the domain of g, in particular, when they are both total, which again shows why we prefer to work in a semantics with total functions only.

Of course, we are not quite done yet. We then have to somehow simplify p ◁ ∘ inits ▪ (a : lp x) to something more efficient. Before we move on, however, proving Proposition 1 turns out to be an interesting challenge in itself.

Intuition

What does Proposition 1 actually say? Let x = [1,2,3] and a = 0. On the left-hand side, we are performing p ◁ on

[] [0] [0,1] [0,1,2] [0,1,2,3]

The right hand side says that we may first filter the tails of [1,2,3]:

[] [1] [1,2] [1,2,3]

Assuming that only [] and [1,2] get chosen. We may then keep the longest prefix [1,2] only, generate all its prefixes (which would be [] [1] [1,2]), and filter the latter again. In other words, we lost no information dropping [1,2,3] if it fails predicate p, since by suffix-closure, p ([0] ⧺ [1,2,3]) ⇒ p [1,2,3]. If [1,2,3] doesn’t pass p, p [0,1,2,3] cannot be true either.

Zantema has a nice and brief proof of Proposition 1 by contradiction. However, the theme of this course has mainly focused on proof by induction and, to keep the possibility of someday encoding our derivations in tools like Coq or Agda, we would like to have a constructive proof.

So, is it possible to prove Proposition 1 in a constructive manner?

The Proof

I managed to come up with a proof. I’d be happy to know if there is a better way, however.